Find the derivative of the expression and simplify fully.

[1irishman]

Homework Statement

k(x) = x^2(2x^3+x)^1/3

Homework Equations

I used the product rule and the chain rule to get to the point that i did.

The Attempt at a Solution

dy/dx(x^2)(2x^3+x)^1/3 + dy/dx(2x^3+x)^1/3(x^2)
= (2x)(2x^3+x)^1/3 + 1/3(2x^3+x)^-2/3(6x^2+1)(x^2) I can't figure out how to solve further.

 

[timthereaper]

A couple of questions:

1) You state k(x) =... and then end up with dy/dx. Is it supposed to be y(x) at the beginning?

2) Why are you doing this: "dy/dx(x^2)(2x^3+x)^1/3 + dy/dx(2x^3+x)^1/3(x^2)"? There's no reason to add them together.

Your solution is correct as far as I've been able to check out. This is how I would rewrite what you have there: dy/dx = (2x)(2x^3+x)^1/3 + 1/3(2x^3+x)^-2/3(6x^2+1)(x^2)

You can also check your solution with wolframalpha.com.

 

[1irishman]

A couple of questions:

1) You state k(x) =... and then end up with dy/dx. Is it supposed to be y(x) at the beginning? it is suppossed to be function named k.

2) Why are you doing this: "dy/dx(x^2)(2x^3+x)^1/3 + dy/dx(2x^3+x)^1/3(x^2)"? There's no reason to add them together. I was using dy/dx to indicate that i was taking the derivative of the first term in the function times the second term in the function and then adding the derivative of the second term of the function times the first term in the function.
I thought the product rule involves adding, no?

Your solution is correct as far as I've been able to check out. This is how I would rewrite what you have there: dy/dx = (2x)(2x^3+x)^1/3 + 1/3(2x^3+x)^-2/3(6x^2+1)(x^2)

You can also check your solution with wolframalpha.com.

The solution is in my book, but i just don't understand how they got it.

 

[1MileCrash]

I don't know what you were adding at the beginning, but I agree with your answer.

Dx[x^2(2x^3+x)^\frac{1}{3}]
=

(x^2)Dx[(2x^2+x)^\frac{1}{3}] + (2x)(2x^3+x)^\frac{1}{3}

..

Dx[(2x^2+x)^\frac{1}{3}]

let 1(x) = x^\frac{1}{3}
let 2(x) = u = (2x^3+x)

Dx[mess] = 1'(u)2'(x)
=
\frac{1}{3}(2x^3+x)^-{\frac{2}{3}}(6x^2+1)

..
Dx[x^2(2x^3+x)^\frac{1}{3}]
=
(x^2)\frac{1}{3}(2x^3+x)^-\frac{2}{3}(6x^2+1) + (2x)(2x^3+x)^\frac{1}{3}

=

(\frac{x^2(6x+1)}{3(2x^3+x)^\frac{2}{3}})+(2x)(2x^3+x)^\frac{1}{3}

That's just how I'd write it, fewer terms and no negative exponents is my preference.

 

[1irishman]

I don't know what you were adding at the beginning, but I agree with your answer.

Dx[x^2(2x^3+x)^\frac{1}{3}]
=

(x^2)Dx[(2x^2+x)^\frac{1}{3}] + (2x)(2x^3+x)^\frac{1}{3}

..

Dx[(2x^2+x)^\frac{1}{3}]

let 1(x) = x^\frac{1}{3}
let 2(x) = u = (2x^3+x)

Dx[mess] = 1'(u)2'(x)
=
\frac{1}{3}(2x^3+x)^-{\frac{2}{3}}(6x^2+1)

..
Dx[x^2(2x^3+x)^\frac{1}{3}]
=
(x^2)\frac{1}{3}(2x^3+x)^-\frac{2}{3}(6x^2+1) + (2x)(2x^3+x)^\frac{1}{3}

=

(\frac{x^2}{3(2x^3+x)^\frac{2}{3}})(6x+1)+(2x)(2x^3+x)^\frac{1}{3}

That's just how I'd write it, fewer terms and no negative exponents is my preference.

in your last line, is that not supposed to be (6x^2 + 1)?

 

[1MileCrash]

Yes! Sorry.

 

[Mark44]

Homework Statement

k(x) = x^2(2x^3+x)^1/3

Homework Equations

I used the product rule and the chain rule to get to the point that i did.

The Attempt at a Solution

dy/dx(x^2)(2x^3+x)^1/3 + dy/dx(2x^3+x)^1/3(x^2)
= (2x)(2x^3+x)^1/3 + 1/3(2x^3+x)^-2/3(6x^2+1)(x^2) I can't figure out how to solve further.

As already pointed out, since you are asked to differentiate k(x), your work should start with k'(x).
Not mentioned is that you are using dy/dx improperly. dy/dx is the derivative of y with respect to x. To indicate that you are about to differentiate with respect to x, use the operator d/dx.

Here is the corrected version of what you show above.

k'(x) = d/dx[(x^2)(2x^3+x)^(1/3)] + [STRIKE]dy/dx(2x^3+x)^1/3(x^2)[/STRIKE]
= (2x)(2x^3+x)^(1/3) + 1/3(2x^3+x)^(-2/3)(6x^2+1)(x^2)

There is some additional work that can be done to put this in a more useful form, using factoring, but all the hard work (i.e., calculus) has been done.

 

[1irishman]

Homework Statement

k(x) = x^2(2x^3+x)^1/3

Homework Equations

I used the product rule and the chain rule to get to the point that i did.

The Attempt at a Solution

dy/dx(x^2)(2x^3+x)^1/3 + dy/dx(2x^3+x)^1/3(x^2)
= (2x)(2x^3+x)^1/3 + 1/3(2x^3+x)^-2/3(6x^2+1)(x^2) I can't figure out how to solve
further.

okay, i came to a fully simplified solution starting right after the above line, here it is:

= 2x(2x^3+x)^1/3 + 1/3(1/(2x^3+x)^2/3(6x^2+1)(x^2)
= 2x(2x^3+x)^1/3 + 6x^4 + x^2/3(2x^3 + x)^2/3
= 3(2x^3 + x)^2/3 (2x)(2x^3 + x)^1/3/3(2x^3 + x)^2/3 + 6x^4 + x^2/3(2x^3 + x)^2/3
= 6x(2x^3 + x) + 6x^4 + x^2/3(2x^3+ x)^2/3
= 12x^4 + 6x^2 + 6x^4 + x^2/3(2x^3 + x)^2/3
= 18x^4 + 7x^2/3(2x^3+x)^2/3
= 1/3x^2(18x^2 + 7)(2x^3 + x)^-2/3
It said to simplify fully so that is what i did.

 

[1irishman]

As already pointed out, since you are asked to differentiate k(x), your work should start with k'(x).
Not mentioned is that you are using dy/dx improperly. dy/dx is the derivative of y with respect to x. To indicate that you are about to differentiate with respect to x, use the operator d/dx.

Here is the corrected version of what you show above.

k'(x) = d/dx[(x^2)(2x^3+x)^(1/3)] + [STRIKE]dy/dx(2x^3+x)^1/3(x^2)[/STRIKE]
= (2x)(2x^3+x)^(1/3) + 1/3(2x^3+x)^(-2/3)(6x^2+1)(x^2)

There is some additional work that can be done to put this in a more useful form, using factoring, but all the hard work (i.e., calculus) has been done.

thank you.

 

[Mark44]

You're welcome!